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If light is hitting a Parabolic Trough defined by $y=x^2$ at a 60 degree angle from vertical so that the effective cross-section of the modified parabola is paramaterized by: x=t, y=t^2, z=tcot(60). Then how would parallel light rays interact with this new paramaterized shape? Would there be a focus? What methods might be employed to investigate this query?

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[edit] Thanks to Juan Sebastian Lozano Muñoz for pointing out that it is necessary to "assume that all light to be parallel to the cross section." enter image description here [edit 2]: Thanks to Geoffrey I can easily understand how this works at the vertex, however it is more difficult to visualize/draw at other points, which involve one vertical and two horizontal components. One horizontal component is a result of the incident light rays incoming at angle $\theta$. The other horizontal component is induced by the parametric parabola described by the equations mentioned above. Therefore it would seem possible to remove the first horizontal component using a theoretical trough defined by the parametric equations x=t, y=t^2, z=tcot(60) instead of $y=x^2$. This would need to assume the incident light rays are at angle $\theta = 0$ from vertical in all directions, which is equivalent to solar tracking.

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So in the above picture the of the "Parametric plot" the light-rays would be parallel to the y-axis. This 2D problem seems much more manageable. Geoffrey was saying that the focal length remains unchanged after accounting for the horizontal distance due to $\theta$, which I have attempted to remove from the equation. The focal length of $y=x^2$ is $y=\frac{x^2}{4f}$ where $f$ is the focal length. Here $f=\frac 14$. Therefore the focal length of the new curve $f_p=\frac{\frac 14}{cos(\theta)}=\frac{\frac 14}{cos(60)}=\frac{\frac 14}{\frac 12}=\frac 12$ Is there a way to prove this? geometrically and/or more rigorously?

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    $\begingroup$ If one considered parallel light rays (or photons) that lied on the plane $C$, then if that plane were perpendicular to the $z$ axis, then there would be a focus. While I have no proof, I suspect that this is false if the plane $C$ doesn't lie perpendicular to the $z$ axis. $\endgroup$ – Juan Sebastian Lozano Nov 6 '13 at 2:34
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    $\begingroup$ Actually... if one assumes all light to be parallel to the cross section then there would be a focus described by a linear equation. Otherwise, there is nothing to say that particles couldn't hit it at an odd angle as to deflect away from the focus line. $\endgroup$ – Juan Sebastian Lozano Nov 6 '13 at 2:49
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    $\begingroup$ @JuanSebastianLozanoMuñoz That is an important assumption that I neglected to include in the question. I will now edit it in. $\endgroup$ – User3910 Nov 6 '13 at 3:29
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For a field of parallel light rays, there will still be a focus if the angle down from vertical is dropped along the axis of symmetry for the trough. The focus will be at the same height as the focus for vertical, parallel rays but shifted over a distance $d$ given by $$ d=f\cot(90-\theta) $$

where $f$ is the focal length of the parabola and $\theta$ is the angle down from the vertical.

In essence, the general properties of the mirror are unchanged other than being translated. This is true because the mirror is uniform along the axis of symmetry so the projection of the ray along the vertical is still reflected the same as it ever was, while the projection along the axis of symmetry is simply rotated $90^o$ one way or the other depending on where it hits the mirror.

If the angle down from vertical has a component that is not along the axis of symmetry, then a portion of the mirror will not receive any light at all while the portion receiving direct light will no longer be at the foot of the trough. By inspection, it can be readily seen that the focus disperses quite quickly.

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  • $\begingroup$ Thankyou for your excellent answer! I have modified the question to make it a little harder. $\endgroup$ – User3910 Nov 6 '13 at 21:28

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